A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes
A. \[\dfrac{L}{2}\]
B. \[L\]
C. \[\dfrac{L}{3}\]
D. \[\dfrac{L}{4}\]

Hint:Before we start addressing the problem, we need to know about angular momentum. It is defined as the measure of the rotational momentum of the rotating body which is equal to the product of the angular velocity of the system and the moment of the inertia to the axis and it is a vector quantity.

Formula used:
The expression of angular momentum is,
\[L = I\omega \]
Where, $I$ is the moment of inertia and $\omega$ is the angular velocity.

Complete step by step solution:
We know that the angular momentum L of a rotating body is given by, \[L = I\omega \]
Now, the angular velocity \[\omega \] can be written as,
\[\omega = \dfrac{{2\pi }}{T}\]
\[ \Rightarrow \omega = 2\pi f\] (f is the frequency of rotation)
Therefore, the momentum L can be written as,
\[L = I\left( {2\pi f} \right)\]
From this we can say that the angular momentum is directly proportional to the frequency of rotation f.

If the new frequency of rotation is half of the initial frequency of rotation, then it can be written as,
\[{f^1} = \dfrac{1}{2}f\]
Then, if we write the angular momentum in terms of ratios we get,
\[\dfrac{{{L^1}}}{L} = \dfrac{{{f^1}}}{f}\]
\[ \Rightarrow \dfrac{{{L^1}}}{L} = \dfrac{{\left( {\dfrac{f}{2}} \right)}}{f}\]
\[ \Rightarrow {L^1} = \dfrac{L}{2}\]
Therefore, if the frequency of rotation is halved, then the angular momentum becomes \[\dfrac{L}{2}\].

Hence, option A is the correct answer.

Note: Angular momentum represents the product of a rotational velocity and inertia about a particular axis of an object. That is, the angular momentum L is proportional to angular speed ω and moment of inertia I and is measured in radians per second.
If there are no net external forces acting on a body then the linear momentum is conserved, when the net torque is zero then the angular momentum is constant or conserved.